Let $f$ be a vector-valued function defined by $f(t)=(e^{-t},\cos(t))$. Find $f$ 's second derivative $f''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(e^{-t},\cos(t))$ (Choice B) B $(-e^{-t},-\cos(t))$ (Choice C) C $(-e^{-t},-\sin(t))$ (Choice D) D $(e^{-t},-\cos(t))$
Answer: We are asked to find the second derivative of $f$. This means we need to differentiate $f$ twice. In other words, we differentiate $f$ once to find $f'$, and then differentiate $f'$ (which is a vector-valued function as well) to find $f''$. Recall that $f(t)=(e^{-t},\cos(t))$. Therefore, $f'(t)=(-e^{-t},-\sin(t))$. Now let's differentiate $f'(t)=(-e^{-t},-\sin(t))$ to find $f''$. $f''(t)=(e^{-t},-\cos(t))$ In conclusion, $f''(t)=(e^{-t},-\cos(t))$.